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- Thread starter darongkun
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- Joined
- Apr 22, 2015

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That's true, and your exercise uses the latter. (It's called 'function notation'.)But cos (pi^2) not equal ( cos pi )^2

\(\displaystyle \frac{1}{\cos(m\pi)^2} = \frac{1}{\cos(m\pi) \cdot \cos(m\pi)}\)

Evaluate factors [imath]\frac{1}{\cos(m\pi)^2}[/imath] for m=1, m=2, m=3 and m=4.

\(\displaystyle \frac{1}{\cos(\pi)^2}\)

\(\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2}\)

\(\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2} \times \frac{1}{\cos(3\pi)^2}\)

\(\displaystyle \frac{1}{\cos(\pi)^2} \times \frac{1}{\cos(2\pi)^2} \times \frac{1}{\cos(3\pi)^2} \times \frac{1}{\cos(4\pi)^2}\)

How does the aggregate product change, as m increases? Using that pattern, extrapolate to m=1000.

- Joined
- Nov 12, 2017

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Is that just your uninformed opinion which you are setting against those of more experienced people, or were you told this?This is my understanding.

The way I read it, [imath]\cos(\pi x)[/imath] is function notation, and must be evaluated before applying the exponent. So [imath]\cos(\pi x)^2[/imath] means [imath][\cos(\pi x)]^2[/imath]. It would have been better if they had written the latter explicitly, but I think anyone with experience will take it this way.

In addition, if you interpret it this way, the problem becomes easily solvable! Otherwise it is not. Sometimes what you need to do is to try one interpretation (especially if someone with knowledge tells you to) and see what happens, then try another to see how it compares. Have you at least